Considering an ordered table with 11 columns:timespan, val1, val2, val3, val4, val5, val6, val7, val8, val9 and val10
Suppose a set of records like:
timespan val1 val2 val3 val4 val5 val6 val7 val8 val9 val10
10/09/2011 0 0 60 80 40 0 0 40 80 0
10/10/2011 0 10 90 30 70 50 50 70 30 90
10/11/2011 10 0 20 0 0 60 60 0 0 20
I need a SQL query (for SQL Server 2012) which returns the last (in time) non zero values for all columns, val1, val2, ..., that is,
val1 val2 val3 val4 val5 val6 val7 val8 val9 val10
10 10 20 30 70 60 60 70 30 20
A similar question can be found at Subquery: how to retrieve the last non-zero value from a column? but it only works for one column and the generalisation to include more columns (as in this case) seems not practical.
You can use first_value()
:
select distinct first_value(val1) over (order by sign(val1) desc, timespan desc) as val1,
first_value(val2) over (order by sign(val2) desc, timespan desc) as val2,
. . .
from t;
Generally, I am opposed to using select distinct
as a replacement for an aggregation query. Unfortunately, SQL Server supports first_value()
as a window function but does not provide an equivalent for aggregation.
Note: the sign()
function is used to put the zero values last. If you can have negative values, then use abs(sign())
.
Another option is a quick unpivot followed by a pivot
Example
Select *
From (
Select top 1 with ties item,value
From YourTable
UnPivot ( Value for Item in (val1,val2,val3,val4,val5,val6,val7,val8,val9,val10) ) u
Where value<>0
Order by Row_Number() over (Partition By item Order by timespan desc)
) src
Pivot (max(value) For item in (val1,val2,val3,val4,val5,val6,val7,val8,val9,val10) ) p
Returns
val1 val2 val3 val4 val5 val6 val7 val8 val9 val10
10 10 20 30 70 60 60 70 30 20
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